= Solution
Let $a=\mu(A)$, $b=\mu(B)$ and $c=\mu(A+B)$. The sum is also symmetric negative definite. Part (a) gives the exponential <norm> bounds, while the <triangle inequality> and submultiplicativity give
$$
\|[e^{xA},B]\|_2\leq\|e^{xA}B\|_2+\|Be^{xA}\|_2
\leq2\|B\|_2e^{xa}.
$$
Applying these estimates to the <exponential product defect identity> yields
$$
\|\Phi(t)-e^{t(A+B)}\|_2
\leq2\|B\|_2\int_0^t e^{(t-x)c}e^{x(a+b)}\,dx.
$$
If $c<a+b$, evaluate the scalar integral to obtain the <decaying bound for symmetric negative matrix splitting>:
$$
\boxed{\|\Phi(t)-e^{t(A+B)}\|_2
\leq2\|B\|_2\frac{e^{t(a+b)}-e^{tc}}{a+b-c}.}
$$
The sign condition is natural: for every unit vector $v$, $v^T(A+B)v\leq a+b$, so the <Rayleigh quotient> gives $c\leq a+b$.
When $c=a+b$, the integrand equals $e^{tc}$ independently of $x$, giving
$$
\boxed{\|\Phi(t)-e^{t(A+B)}\|_2\leq2t\|B\|_2e^{tc}.}
$$
This also follows by taking the continuous limit of the quotient as $a+b-c\to0$. There is no division-by-zero exception in the original integral argument. All exponents are negative, so these bounds decay for large $t$; they are coarse bounds and need not reproduce the sharper quadratic small-$t$ error.
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